# Real Analysis

Open and Closed Sets

# Limit Points

Say \(p \in X\) is a *limit point* of \(E \subset X\) if every neighborhood of \(p\) (i.e. \(N_r(p)\) for all \(r\)) contains a point \(q \neq p\) such that \(q \in E\).

So, a point \(p\) is not a limit point of \(E\) if there exists a neighborhood \(N\) such that \(N\) does not contain any other points of \(E\).

For example,

- In \(\mathbb{R}\), \(G = \{\frac{1}{n} : n \in \mathbb{Z}\}\). \(0\) is a limit point of \(G\). In \(\mathbb{R}\), a neighborhood of a point \(p\) is an open interval \((p - \epsilon, p + \epsilon)\).

# Isolated Points

A point \(p \in X\) is an *isolated point* of \(E \subset X\) if \(p \in E\) and \(p\) is not a limit point of \(E\).

For exmple,

- All points of \(G = \{\frac{1}{n} : n \in \mathbb{Z}\}\) are isolated.

# Interior Points

A point \(p \in X\) is an *interior point* of \(E \subset X\) if \(p \in E\) and there exists a neighborhood \(N\) of \(p\) such that \(N \subset E\).

For example,

- \(G = \{\frac{1}{n} : n \in \mathbb{Z}\}\) has no interior points.
- In \(\mathbb{R}\), \(\emptyset\) has no limit points, no interior points, and no isolated points, for \(\mathbb{R}\) has no isolated points but all points in \(\mathbb{R}\) are interior and are limit points, and \(\mathbb{Q}\) has no isolated points and no interior points but all points in \(\mathbb{Q}\) are limit points.
- In (\(\mathbb{R}\), \(\text{discrete}\)), \(\emptyset\) has no limit points, no interior points, and no isolated points, \(\mathbb{R}\) has no limit points but all points in \(\mathbb{R}\) are interior and isolated, and \(\mathbb{Q}\) has no limit points but all points in \(\mathbb{Q}\) are interior and isolated.

**Theorem.** If \(p\) is a limit point of \(E\), then every neighborhood of \(p\) contains infinitely many points of \(E\).

*Proof (by contradiction).* Assume there exists a neighborhood \(N\) of \(p\) with only finitely many points in \(E\), call them \(e_1, e_2, \dots, e_n\).

Let \(r = \min_{i = 1}^{n} \{d(p, e_i)\}\) (\(r\) exists because \(e_i\) is finite). But \(N_r(p)\) has no points other than \(p\), a contradiction! QED.

# Open Sets

A set \(E\) is *open* if every point of \(E\) is an interior point of \(E\).

For example,

- In \(\mathbb{R}\), any open interval \((a, b)\) is open.
- In \(\mathbb{R}\), \(\emptyset\) is open.
- In \(\mathbb{R}\), \(\mathbb{R}\) is open.

**Lemma.** Neighborhoods are open.

*Proof.* Consider a neighborhood \(N\) around a point \(p\) with radius \(r\). Consider a point \(q \in N\). So \(d(p, q) < r\). Let \(a = d(p, q)\). Let \(r' = r - a > 0\). Consider \(N_{r'}(q)\). For all points \(x \in N'\), we have \(d(q, x) < r'\). Then \(d(x, p) \leq d(x, q) + d(q, p) < r' + a = r.\) Thus, \(x \in N\). Therefore, \(N_{r'} \subset N_r\). QED.

# Closed Sets

A set \(E\) is *closed* if \(E\) contains all its limit points.

For example,

- In \(\mathbb{R}\), a single point \(p\) is closed.
- In \(\mathbb{R}\), \(\emptyset\) is closed.
- In \(\mathbb{R}\), \(\mathbb{R}\) is closed.
- In \(\mathbb{R}\), any closed interval \([a, b]\) is closed.
- In \(\mathbb{R}\), any half-open interval \([a, b)\) or \((a, b]\) is neither open nor closed.

Sets which are both open and closed is called *clopen*.

# Closing a Non-Closed Set

The *closure* of \(E\), denoted \(\overline{E}\), is the union of \(E\) and the set of its limit points, denoted \(E'\). In other words, \(\overline{E} = E \cup E'\).

**Theorem.** \(\overline{A}\) is a closed set.

*Proof.* To prove \(\overline{A}\) is closed, we need to prove all limit points \(p\) of \(\overline{A}\) is in \(\overline{A}\). In order to prove that, we show that for all limit points \(p\) of \(\overline{A}\), \(p\) is a limit point of \(A\), i.e. \(p \in A' \subset \overline{A}\).

For all \(r > 0\), consider the neighborhood \(N\) around \(p\) with radius \(r\), i.e. consider \(N_r(p)\). We need to show that there exists a point \(q \in N \cap A\). Since \(p\) is a limit point of \(\overline{A}\), there exists a point \(t \in N \cap \overline{A}\). Thus, \(t\) is in \(A\) or \(t\) is a limit point of \(A\). When \(t \in A\), we have \(q = t \in A\). When \(t\) is a limit point of \(A\), consider a neighborhood \(M\) around \(t\) such that \(M \subset N\) (\(M\) must exist because \(N\) is an open set and \(t\) is in \(N\), which makes \(t\) an interior point of \(N\)). Since \(t\) is a limit point of \(A\), there exists \(q \in M \cap A\), which implies that \(q \in N \cap A\), as desired. QED.

**Theorem.** A set \(A\) is closed if and only if \(A = \overline{A}\).

*Proof.* Let \(A'\) be a set containing all limit points of \(A\).

Show that \(A\) is closed only if \(A = \overline{A}\): Since \(A\) is closed, we have \(A' \subset A\). So, \(A \cup A' \subset A\). Hence, \(\overline{A} \subset A\). Since, clearly, \(E \subset \overline{E}\), \(E = E'\).

Show that \(A\) is closed if \(A = \overline{A}\): Suppose \(A = \overline{A}\), which implies \(A\) contains all its limit points. Hence, \(A\) is closed. QED.

**Theorem.** If \(E \subset F\), where \(F\) is closed, then \(\overline{E} \subset F\).

*Proof.* Since \(F\) is a closed set, \(F\) contains all its limit point. In particular, it contains all \(E\)’s limit points. Therefore, \(F\) contains \(\overline{E}\). QED.

\(\overline{E}\) is the smallest closed set containing \(E\).

# Relationship between Open and Closed Sets

For a set \(E\), the complement of \(E\), denoted \(E^c\), is \(X \setminus E = \{x : x \notin E\}\).

**Theorem.** \(E\) is open if and only if \(E^c\) is closed.

*Proof.* \[\begin{align}
E \text{ is open} & \Leftrightarrow \text{all points } p \in E \text{ has a neighborhood } N \subset E \\
& \Leftrightarrow \forall p \in E, \exists N_r(p) \subset E \\
& \Leftrightarrow \forall p \in E, \exists N_r(p) \cap E^c = \emptyset \\
& \Leftrightarrow \forall p \in E, p \text{ is not a limit point of } E^c \\
& \Leftrightarrow E^c \text{ contains all its limit points} \\
& \Leftrightarrow E^c \text{ is closed}
\end{align}\] QED.

For example,

- Let \(K_i = \left[-1 + \frac{1}{n}, 1 - \frac{1}{n}\right]\) (each \(K_i\) is closed). We have \(\bigcup_{n = 1}^\infty K_i = (-1, 1)\), which is open and not closed.
- Let \(K_i = \left(-\frac{1}{n}, \frac{1}{n}\right)\) (each \(K_i\) is open). We have \(\bigcap_{n = 1}^\infty K_i = \{0\}\), which is closed and not open.

**Lemma.** Let \(\{E_\alpha\}_{\alpha \in I}\) be a collection of sets. Then \((\bigcup_{\alpha \in I} E_\alpha)^c = \bigcap_{\alpha \in I} E_\alpha^c\).

*Proof.* \[\begin{align}
x \in LHS & \Leftrightarrow x \notin \text{any } E_\alpha \\
& \Leftrightarrow x \in E_\alpha^c \text{ for all } \alpha \\
& \Leftrightarrow x \in \bigcap_{\alpha \in I} E_\alpha^c
\end{align}\] QED.

**Theorem.** Any arbitrary union of open sets is open.

*Proof.* Let \(\{E_\alpha\}_{\alpha \in I}\) be a collection of open sets. Let \(E = \bigcup_{\alpha \in I} E_\alpha\). Consider any point \(x \in E\). Then, \(x \in E_\beta\) for some \(\beta \in I\). Since \(E_\beta\) is open, there exists a neighborhood \(N\) around \(x\) such that \(N \subset E_\beta\). Hence, \(N \subset E\). Therefore, \(x\) is an interior point of \(E\), as desired. QED.

**Theorem.** Any arbitrary intersection of closed sets is closed.

*Proof.* Let \(\{E_\alpha\}_{\alpha \in I}\) be a collection of closed sets. Consider any such set \(E_\alpha\). Since \(E_\alpha\) is closed, \(E_\alpha^c\) is open. From the previous theorem, we have \(\bigcup_{\alpha \in I} E_\alpha^c\) is open. Hence, \(\left(\bigcup_{\alpha \in I} E_\alpha^c\right)^c\) is closed. Therefore, \(\bigcap_{\alpha \in I} E_\alpha\) is closed. QED.

**Theorem.** Any finite intersection of open sets is open.

*Proof.* Let \(E_1, E_2, \dots, E_n\) be open sets. Let \(E = \bigcap_{i = 1}^n E_i\). Consider any point \(x \in E\). So, \(x \in E_i\), for all \(1 \leq i \leq n\). For each \(E_i\), there exists a neighborhood \(N_{r_i}(x)\) such that \(N_{r_i} \subset E_i\), since \(E_i\) is an open set. Let \(r = \min\{r_1, r_2, \dots, r_n\}\). Then \(N_r(x)\) is contained in \(E\), as desired. QED.

**Theorem.** Any finite union of closed sets is closed.

*Proof.* Let \(E_1^c, E_2^c, \dots, E_n^c\) be closed sets. Then, \(E_1^c, E_2^c, \dots, E_n^c\) are open sets. Hence, \(\bigcap_{n = 1}^n E_i^c\) is open. So, \(\left(\bigcap_{n = 1}^n E_i^c\right)^c\) is closed. Thus, \(\bigcup_{n = 1}^n E_i\) is closed. QED.

# Dense Sets

A set \(E\) is dense in a metric space \(X\) if every point of \(X\) is a limit point in \(E\) or in \(E\). Or, equivalently, \(\overline{E} = X\). Also, equivalently, every open set of \(X\) contains a point in \(E\).

For example,

- \(\mathbb{Q}\) is dense in \(\mathbb{R}\).
- Can periodic functions be approximated by sums of sines and cosines (Fourier analysis)? Or, equivalently, are linear combinations of sines and cosines dense in a set of all periodic functions? More generally, is a certain subset of functions dense in a whole space of functions?